energy (heat) may be expressed in joules or calories. 1 calorie (cal) = 4.184 joules (j) how many...
TRANSCRIPT
CALCULATING ENERGY CHANGES
HEATING CURVE OF WATER
COOLING CURVE OF WATER
PHASE DIAGRAM
ENERGY CONVERSIONS
Energy (heat) may be expressed in joules or calories.
1 calorie (cal) = 4.184 joules (J)
How many joules in 60.1 calories?
How many calories in 28.4 J?
CALCULATING ENERGY CHANGES
How much energy is required to heat 25 g of liquid water from 25oC to 100oC? Specific heat capacity of liquid water is 4.18 J/goC.
Q = mcΔT Q (energy) = m(mass) x c(specific heat) x
ΔT (change in temperature = Tf – Ti) Q = (25 g)(4.18 J/goC)(100o – 25oC) If ΔT is positive, Q will be positive. If ΔT is
negative, Q will be negative.
How many joules of heat are given off when 5.0 g of water cools from 75oC to 25oc? (Specific heat of water = 4.18 J/goC)
YOUR ASSIGNMENT: DUE MON.3/29
Page 329 Practice Problem 10.2 Page 330 Practice Problem 10.3 Page 333 Section 10.2 Review
Questions 2-6
ENERGY DURING PHASE CHANGE
Energy is increasing/decreasing during phase changes even though temp. remains constant.
Q = mass(m) x heat of fusion (Hf) for melting and freezing
Q = mass(m) x heat of vaporization (Hv) for boiling and condensing
Values will be positive for melting and boiling Values will be negative for freezing and
condensing
How many calories are given off when 85 g of steam condense to liquid water? (Hv= 539.4 cal/g)
How many joules does it take to melt 35g of ice at 0oC? (Hf = 333 J/g)
How many joules are required to convert 10.0g of ice at -10.0oC to steam at 150. oC?
Given: Specific heat of Specific heat of water = 4.184 J/g C° Specific heat of Steam = 2.03 J/g C° Specific heat of ice = 2.06 J/g C° Hv water = 539.4 cal/g Hf water = 333.0 J/g
5 part calculation
YOUR ASSIGNMENT: DUE TUES. 3/30
Page 497 Practice Problem 14.2 Page 497 Section 14.1 Review 1, 3, 4-7 Page 503 Section 14.2 Review 1, 4, 6